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Exact value of convergence

  1. Aug 11, 2008 #1
    Hi there!

    I 've been thinking about finding the exact value series converge to. How does one do this? I know the convergence proof methods (minor, grater, Leibniz's, quotien, square-root and Cauchy's criteria) - but they just prove the convergence itself and do not give us the exact value.

    Here's an example if what I'm talking about

    [tex]\displaystyle{\int}e^{-x^2}dx=?[/tex]

    [tex]e^x=\displaystyle{\sum_{n=0}^{\infty}}\frac{x^n}{n!}[/tex]

    so

    [tex]\displaystyle{\int}e^{-x^2} dx=\displaystyle{\int}\displaystyle{\sum_{n=0}^{\infty}}\frac{(-x^2)^n}{n!}dx=\displaystyle{\sum_{n=0}^{\infty}\int}\frac{(-1)^nx^{2n}}{n!}dx[/tex]

    [tex]\displaystyle{\int}e^{-x^2} dx=C+\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}[/tex]

    Now this should be the series definition of the Gaussian error function. Suppose C=0 and we're going to take the normalised function for simplicity. How can the following two limits be calculated:

    [tex]\displaystyle{\lim_{x\rightarrow +\infty}}\frac{2}{\sqrt{\pi}}\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}=?(answer: 1)[/tex]

    [tex]\displaystyle{\lim_{x\rightarrow -\infty}}\frac{2}{\sqrt{\pi}}\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}=?(answer: -1)[/tex]

    The answers I have seen from Wikipedia: Error function - Wikipedia, the free encyclopedia (cf. the graph)

    Thanks in advance, Marine!
     
  2. jcsd
  3. Aug 11, 2008 #2

    arildno

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    As you can see, Marin, the series definition about 0 is wholly impenetrable with respect to deducing those answers!

    You aren't any less clever than me, I don't see out of that series definition why it must equal 1, as little as you do.

    So, can you think why I nonetheless DO know that it equals 1?
     
  4. Aug 11, 2008 #3
    One way to calculate this limit is like this:

    [tex]
    \lim_{x\to\infty} \frac{2}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)n!}
    = \lim_{x\to\infty}\frac{2}{\sqrt{\pi}} \int\limits_0^x\Big(\sum_{n=0}^{\infty}\frac{(-1)^n u^{2n}}{n!}\Big) du
    = \lim_{x\to\infty}\frac{2}{\sqrt{\pi}} \int\limits_0^x e^{-u^2} du = 1
    [/tex]

    where the knowledge

    [tex]
    \int\limits_0^{\infty} e^{-u^2}du = \frac{\sqrt{\pi}}{2}
    [/tex]

    was used in the last step. :wink: :biggrin: :rofl:
     
  5. Aug 11, 2008 #4
    I'm not sure if I was being funny, but hey, I'm serious with this now:

    If you have some series representation for some function, like

    [tex]
    f(x) = \sum_{n=0}^{\infty} f_n(x),
    [/tex]

    and you want to calculate

    [tex]
    \lim_{x\to \infty} f(x),
    [/tex]

    then if the limits

    [tex]
    \lim_{x\to\infty} f_n(x)
    [/tex]

    exist, it can be possible to calculate the limit like

    [tex]
    \lim_{x\to \infty} \sum_{n=0}^{\infty} f_n(x) = \sum_{n=0}^{\infty} \lim_{x\to\infty} f_n(x).
    [/tex]
    (it can also be, that this does not work, though)

    However, if the limits

    [tex]
    \lim_{x\to\infty} f_n(x)
    [/tex]

    are already diverging, then there is no chance of getting anything out by changing the order of limit and summation, and then this series representation alone is useless for the purpose of calculating the limit of the original problem. The only way to calculate the limit, is to find some other information about the function f(x).
     
  6. Aug 11, 2008 #5
    arildno, the answers, as I wrote, are deduced from the graph of the function in Wikipedia :) I was wondering how to get to them analytically :)

    jostpuur, this is a nice method, thanks a lot I liked it very much! But can it be applied to all types of series?

    e.g.: [tex]\displaystyle{\sum_{n=1}^{+\infty}}\frac{(-1)^{n-1}}{(n+1)a^{2n}}[/tex]

    The series is said to converge absolutely for [tex]lal>0[/tex] but how do we know what value it converges to? Does your method help also here? (btw, what's the Latex command for absolute value?)
     
  7. Aug 11, 2008 #6
    Uh oh. I was actually fooling around a little bit only. Basically you did the effort of calculating Taylor series of the e^(-x^2) function, and its integral function, and then I just substituted these functions back into their Taylor series! I didn't do much there.

    If you are given a function, and you calculate its Taylor series, it is then simple thing to substitute the function to anywhere where you see that Taylor series.

    But if you are given series without knowing from which function it has been calculated with the Taylor's formula, then you are in trouble. It is more difficult to start finding the function this way.


    Integration may involve special trickery too. If you want to look how the Gaussian integration formula is derived (the formula I used in the last step), check http://en.wikipedia.org/wiki/Gaussian_integral

    Just write |a|>0. -> [tex]|a|>0[/tex]. The character | comes somehow from the keyboard, but I'm not sure if its the same all over the world.
     
  8. Aug 11, 2008 #7
    This is the same thing as

    [tex]
    -\sum_{n=1}^{\infty}\frac{1}{n+1}\Big(-\frac{1}{a^2}\Big)^n,
    [/tex]

    so the problem is to calculate series

    [tex]
    \sum_{n=1}^{\infty} \frac{1}{n+1}x^n.
    [/tex]

    Once this series is calculated, it is simple thing to substitute [tex]x=-1/a^2[/tex]. If I now looked this right, this series can be calculated by substituting

    [tex]
    \frac{1}{n+1}x^n = \frac{1}{x}\int\limits_0^x u^n du.
    [/tex]

    When you change the order of integration and summation, you end up with geometric series.
     
    Last edited: Aug 11, 2008
  9. Aug 11, 2008 #8

    arildno

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    Okay, here is how it goes (take a deep breath!)

    1. Defining a definite integral:
    We define I as:
    [tex]I=\int_{-\infty}^{\infty}e^{-x^{2}}dx[/tex]

    2. Squaring I, and using Fubini's theorem:
    We have:
    [tex]I^{2}=(\int_{-\infty}^{\infty}e^{-x^{2}}dx)^{2}=\int_{-\infty}^{\infty}e^{-x^{2}}dx*\int_{-\infty}^{\infty}e^{-x^{2}}dx=\int_{-\infty}^{\infty}e^{-x^{2}}dx*
    \int_{-\infty}^{\infty}e^{-y^{2}}dy[/tex]
    since changing the name of a dummy variable from x to y doesn't mean anything.

    But now, we may invoke Fubini's theorem:
    [tex]I^{2}=\int_{-\infty}^{\infty}e^{-x^{2}}dx*
    \int_{-\infty}^{\infty}e^{-y^{2}}dy=\int_{-\infty}^{\infty}\int_{-infty}^{\infty}e^{-x^{2}}e^{-y^{2}}dxdy=\int_{-\infty}^{\infty}\int_{-infty}^{\infty}e^{-(x^{2}+y^{2})}dxdy[/tex]
    where Fubini allows us to convert a product of two single-variable integrals into a double integral over their product integrand.

    3. Switch to polar coordinates:
    [tex]I^{2}=\int_{-\infty}^{\infty}\int_{-infty}^{\infty}e^{-(x^{2}+y^{2})}dxdy=\int_{0}^{2\pi}\int_{0}^^{\infty}e^{-r^{2}}rdrd\theta=\pi[/tex]
    where the integration is trivial.

    4. Summing up:
    We therefore have, by taking the square root:
    [tex]I=\sqrt{\pi}[/tex]
    You can do the rest.
    :smile:
     
  10. Aug 11, 2008 #9
    Well this is what I meant. See I finished school 2 months ago and started reading some things in preparation for my studies :) It was hard to understand the elementary functions I was taught are a tiny fraction of all functions but in the series I saw a universal method for defining functions.

    Now it even seems to me that it's somehow the origin of it all. If you come upon an unknown function during some calculations you won't see it in a form like sin(x) or ln(x) or arsinh(x) or whatever.. but you'll still have the series expansion.

    That's why I'm asking for all the things you could determine just by having the series and not the function it's derived from. Well, divergence proof tells us maybe a lot but convergence proof seems to me somehow less useful, if we don't know what it converges to.


    When analysing a function I am taught to use the following algorithm:
    1. domain
    2. limits on the verges of the domain, asymptotes and asymptotic curves
    3. symmetry
    4. intersection points with the X- and Y-axis
    5. derivatives (1st, 2nd and 3rd)
    6. calculation of extrema
    7. calculation of inflex points

    But given just the series seems to me very difficult to determine all these things (except maybe symmetry)

    So do you have any ideas how is one supposed to proceed in such cases?

    maybe on the following examples:

    [tex]I. \displaystyle{\sum_{n=0}^{+\infty}}\frac{x^n}{n!}[/tex]

    as if we do not know where the series comes from (so, no e^x)


    [tex]II. \displaystyle{\sum_{n=0}^{+\infty}}\frac{(-x^2)^n}{n!}[/tex]

    without using e^(-x^2)

    **thanks for deriving the Gaussian formula for me. I think double integrals have to wait for their time (cannot understand why it's legal to convert the double integral in to separate single integrals and when one is allowed to change the order of integration)
     
  11. Aug 11, 2008 #10

    arildno

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    It is GREAT that you single out the crucial technical points that HAVE to be proven, Marin!
    Keep that in mind when you encounter the proofs, so that you can see, to your satisfaction, that they are, indeed, valid.
     
  12. Aug 12, 2008 #11
    Any suggestions will be much appreciated!
     
  13. Aug 12, 2008 #12

    arildno

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    Except for 5, which is trivial for (most) power series (just do term-wise differentiation), the other stuff is generally nasty stuff. 1. can also often be done, but may require quite a bit of sophistication.

    The REAL utility of stuff like power series, apart from theoretical stuff (very much so), is that they enable us to CALCULATE, to an arbitrary degree of accuracy what, for example [itex]e^{\pi}[/itex] is in terms of decimals.

    And that can be very useful at times..
     
  14. Aug 12, 2008 #13
    hmmm I read that they are useful for solving DEs. This method is still not to my knowledge. But what if we have to analyse these series? Just put them in a program to do it for you, or how do you proceed? But you still cannot obtain exact values(e.g. for an asymptote) - you can see - 1,23524672 but you'll never know this is sqrt(pi) or something /used just as example/ you could use theoretically...
     
  15. Aug 12, 2008 #14

    arildno

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    I put that under the umbrella of "theoretical stuff". I should have been clearer on that.
     
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